# Full text of "Text Book Of Mechanical Engineering"

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```y

,<\                          1104                            Appendix VI.

(z>2\
H------)
2£'

--       /47T   77/A2

H"li5 V) ^

Next take careful measurements of fly-wheel section to find I
in ins. and Ibs., deducing

I   n 2
Energy of rotation  =  -0001714 — -™H   foot pounds,

(by calculation)                              I2   6o

and the two results ought to agree very closely.

M of Fly Wheel. — Professor Perry has given the name ' M of
a fly-wheel ' to a constant belonging to each particular fly-wheel,
such that MN2 = energy in foot pounds. Hence, if the energy
be divided by N2, the value of M is found. The method is very
convenient, for conversely the energy may be at any time calcu-
lated if the speed of revolution be known. Also

•0001714 ~-2-N2  -  MN2
.*. M  =   '000,001,19!  ==

For a cylinder, M  =   Jfb

y

and for a ring, M

534,800

while the fly-wheel)      ^           /Vr           r*n?

experiment" gives}     M  " ^(H~-oi7

Example 77. — A Fly-wheel is required to store 12,000 foot pounds
of energy as its speed increases from 98 to 102 revs, per m. What is
its I ?

The wheel being a solid disc of cast iron whose thickness is ^ of
its diameter, find D. (Hons. Applied Mechs. Exam.)

Kinetic energy = '0001714 I (N^

*=* *oooi7i4(io22~982)I = 12,000

I = -ze/R2 = ••_J2£B*__ = 8; 5I4 in absolute units
800 x -0001714      -&-±   (feetandlbs0```